It’s A Math, Math World (Wilcoxon Part 2)

The following example is from the textbook, General Statistics, by Chase and Brown (2000).

So far, every hypothesis test we have considered had required an assumption of normality or near-normality.  Sometimes, though, the distribution of the population is non-normal or unknown and we use a nonparametric test.  An easy example of such a test is the Wilcoxon Signed-Rank test for testing the value of a median of a population. This is superior to the Sign Test (see last post) because it does not ignore the magnitude of the data as the Sign Test does. Thus, the Wilcoxon test is more sensitive and will more often reject a false null hypothesis.

A necessary condition for this test is that the data must be continuous. Also, we assume symmetry in the distribution of the population.

Last week we looked at the Wilcoxon Signed-Rank Test for a single population median (see post Wilcoxon Part 1).  Today we will look at the difference between 2 medians (a 2 population case).  In this case, the samples are dependent – that is, the data is obtained in pairs.  Also, the data must be continuous and the 2 distributions must have similar shapes.

Example: A college professor claims that a remedial English course will help students whose English skills are deficient. Twenty-five students who failed a pre-test are given the course and then take a post-test. Is the professor’s claim justified at the 5% level of significance?

The professor claims that the pretest scores (X1) tend to be lower than the scores on the posttest (X2). This means that the values of the difference D= X1-X2 will be less than zero (i.e. Median (differences) < 0). Thus, the hypotheses are:

H0: Median (Differences) = 0

HA: Median (Differences) < 0

Level of significance: α = 0.05

Test statistic: We use W+ as a test statistic. From the table below, we see that W+ = 68.5

Critical Region: From the appropriate table, we see that the critical value for a one-tailed test     with α = 0.05 and n=25 is c=101. Thus the critical region consists of values of W+ ≤ 101.

Conclusion: The observed value of 68.5 is in the critical region, so we reject H0. This appears that the professor’s claim appears to be correct. Scores on the pretest appear to be lower than those on the posttest which suggests that the course is effective.

Note on Tied Ranks: In this case, we have values of |D| that are the same. We have that case in the ranks of the elements in positions 11, 12, 13 and 14. IN the case, we average the ranks and get 12.5 and use the common rank for all 4 elements. The next rank would start at 15 and proceed as usual.

PRE-TEST (X1) POST-TEST (X2) DIFFERENCE    (X1-X2) |D| SIGNED RANK
46 76 -30 30 -25
27 36 -9 9 -7
37 53 -16 16 -12.5
34 55 -21 21 -18
20 12 8 8 6
38 50 -12 12 -10
10 36 -26 26 -22
24 18 6 6 4
20 21 -1 1 -1
39 57 -18 18 -15
16 27 -11 11 -9
20 48 -28 28 -23
47 70 -23 23 -19
45 25 20 20 17
40 50 -10 10 -8
46 39 7 7 5
32 51 -19 19 -16
49 33 16 16 12.5
45 69 -24 24 -20
49 52 -3 3 -2
44 60 -16 16 -12.5
45 20 25 25 21
16 12 4 4 3
41 70 -29 29 -24
48 64 -16 16 -12.5

Case of Zero Differences: If any of the values of D are zero, then we use the following procedure.

If there is an even number of zeros, each zero is assigned the average rank for the set, and then half of them are assigned a plus sign and the other half a minus sign.

Ex.  If there are 4 zeros, then we would assign the ranks 1, 2, 3, and 4 giving us an average rank of 2.5. We would end up with the signed ranks: -2.2, -2.5, 2.5, and 2.5.

If there is an odd number of zeros, we discard one of them, reduce the sample size by 1 and proceed as in the even case.

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